The h Bar

Emilio Pisanty

12:20 AM

@DanielSank you may find these graphics assets useful

cdn.sstatic.net/Sites/quantumcomputing/img/…

the svg sprite sheet has a fully vector version of the logo

(and if the question is "How?!", the answer is Chrome > right click > inspect > Sources.)

I'll try and give that ad a go if I can find a time window

but if I can't find the time, the official logo should help

also, hey, @dmckee how come I gave away 250 rep and I'm back on 7th?

Phase

Lmao

vzn: everyone in conventional physics is wrong

acm: proof?

vnz: everyone is paranoid

classic hbar

0celo7

@Phase omg you're alive

Secret

1:03 AM

@AkivaWeinberger you cannot really break the laws of physics, at least without end up accidentally killing yourself

Fun fact: for any question, it becomes philosophy once the word "exists" was added

Afred

in Mathematics, 2 mins ago, by Niing

Is the reverse of Symmetry of second derivatives, or Schwarz's theorem, also correct?

Iff?

Secret

1:26 AM

Consider the following:

Suppose in a region of space $X$ there is a "Probability Field" $P : X \to [0,1]$. A particle passes through $X$ in some welldefined trajectory $\gamma$.

Therefore the existence and hence the interaction probability of the particle with a stationary one with unity is given by:

$$\frac{1}{||\gamma||}\int P(s)ds$$

(Almost forgot: this is worldbuilding)

Now consider two particles moving together along $\gamma$ but their probability fields are given by $P_1$ and $P_2$ respectively. The probability of interaction is then given by:

dmckee

1:49 AM

@EmilioPisanty Don't know. How often does the user listing update? Maybe it just hasn't tripped yet.

Secret

$$\frac{1}{||\gamma||^2}\int_{\gamma^2}P_1(s)P_2(s)-J_{12}(s)ds $$

where J is the joint probability, due to correlations between the two individual probability distributions

Now to think about how to formulate two particle trajectories that intersect only at some points, cause any integral or sum that gives the interaction probability only depends on the intersections

[Full Scope]

Phase

2:23 AM

@0celo7 sadly

@0celo7 heading to bed rn tho

Secret

(So bored of the silence of h bar)

Let $X$ be a region of space, with positions denoted by $\textbf{x}$. Define on this space a probability density $\rho : X \to [0,1]$.

Let some trajectory $\gamma$ be a continuous set of points in $X$. I.e. $\gamma : [t_0,t_1] \to X$

Thus, $\gamma$ is a function of positions wrt time

For the moment being, assume time is Newtonian. We will consider the relativistic case later

Secret

2:47 AM

We also assume the particle is classical, thus its state of motion is described by the tuple $(\textbf{x},\textbf{v},\rho(t))$

where $\textbf{v} = \dot{\textbf{x}}$

These particles have a probabilistic existence in that given a state, there is a nonzero probability the particle is not present with respect to its external environment, thus two particles whose trajectories crosses, can potentially pass through each other without interactions

All results of one classical particle are the same as their deterministic classical counterparts, as in a system where the particles are isolated and noninteracting, the probability density has no observable consequences

The case becomes more interesting when we consider two interacting particles, thus we will discuss from there

Let $X=\Bbb{R}^2$. Denote the initial conditions of the particles be $((-5,0),(1,0),\rho_1(\textbf{x}_1))$ for particle 1 and $((5,0),(-1,0),\rho_2(\textbf{x}_2))$ for particle 2

typo

Let $X=\Bbb{R}^2$. Denote the initial conditions of the particles be $((-x,0),(v,0),\rho_1(\textbf{x}_1))$ for particle 1 and $((x,0),(-v,0),\rho_2(\textbf{x}_2))$ for particle 2, which after collision, bounces off with an inclined angle of $\theta$

The position of the particles are given as follows:

$\textbf{x}_1(t) = (-x,0)+\int_{0}^{t}(v,0)dt=(-x+vt,0)$
$\textbf{x}_2(t) = (x,0)+\int_{0}^{t}(-v,0)dt=(x-vt,0)$

rob

3:25 AM

12

A: Inverse square rule for strong forces

Most of the forces induced by a point particle follows the 1/r^2 rule No, it's the forces mediated by point particles with no mass and charge that follow the the 1/r^2 rule. then why does strong force don't obey it? The inverse square law is a consequence of the particles having no mas...

I always thought that the non-perturbative nature of the strong force arose both from the non-neutral gluon and from the large strong coupling constant. Am I mistaken, and one of those features is enough? Or should I add an answer talking about the coupling constants?

Secret

Using the above, we can deduce the probability of the particles at any point is given by $\rho_1(-x+vt)$ and $\rho_2(x-vt)$

and hence the probability that the collision will occur at the origin is given by (assuming the two probability distributions are uncorrelated, hence $\rho_{12}=\rho_1\rho_2$)

$\rho_1(0,\frac{x}{v})\rho_2(0,\frac{x}{v})$

vzn

@Phase lol did you memorize your entire QM textbook word-for-word yet? incl every formula? + index? get back to us when you have... because until you have youre not a real physicist™ :P

Secret

3:40 AM

and thus, the probability that the two particles obtain some final velocity depends on the initial energy of the system

which is given by $mv^2 =\frac{1}{2}mv_f^2 \implies v_f = \sqrt{2}v$ and the final velocities of the particles are $(\frac{\sqrt{2}}{2}v\cos \theta,-\frac{\sqrt{2}}{2}v\sin \theta)$ and $(-\frac{\sqrt{2}}{2}v\cos \theta,\frac{\sqrt{2}}{2}v\sin \theta)$ respectively

We will deal with more complicated cases later, such as this:

Claim: The White House is a conspiracy

Phase

4:41 AM

@vnz I would be able to find a reference for what I'm referencing. That's what he was asking you for. You failed to provide any, and reply with crappy responses like that

vzn

@Phase lol huh? who is asking for what? speaking of "crappy responses" when are you going to figure how to spell my id right? is 3 consecutive letters too much?

Phase

acm asked for a reference, and I mistyped the letters because I'm multitasking. If that's the best argument you have then you're more of a joke than I thought.

vzn

@Phase lol ACM doesnt really give a ---- about a ref. gave him one months ago. you probably dont really give a ---- either.

Phase

is the reference reputable?

vzn

@Phase do you even have the slightest )( clue + actual interest wtf we're talking about?

Sir Cumference

4:50 AM

@ACuriousMind So...a tuple of vectors?

Not really sure what you mean by "vector of vector of vectors"

Phase

@vzn "@vzn Where are they written down?" asking about the equations

vzn

@Phase its interesting how you consistently mystype the letters, seems more dyslexic like. ofc feel free to laugh at whatever jokes you find funny.

@Phase lol what eqns?

Phase

Consistently? Are you autistic? Cite more than one example please

Exactly that's his point, you seemed to have none yet claimed they existed.

vzn

@Phase seems like you dont actually want to talk about physics but just wanna play some kind of alphamale games instead.

Phase

No. You just wont accept that you consistently weasel out of any valid points against you. You keep making bad arguments instead, and saying things that are just wrong like that I got your name wrong multiple times, not that that is even relevant to this.

You claim light is rippling space.

Naturally you're asked for proof

you deny to give any and throw a fit

and then act as if it's everyone elses fault

vzn

4:55 AM

@Phase lol not autistic heres 5 :P chat.stackexchange.com/search?q=vnz&room=71

Phase

*2

only 2 are by me, and they werent because I struggle spelling it's because your name isn't memorable.

But again, you don't respond to the real points.

Goodnight.

vzn

@Phase would like to do so at length with anyone who really gives a ---- :) goodnite sleep well

Secret

5:22 AM

@SirCumference a (0,3) tensor acts on a tuple of 3 vectors $(\textbf{v}_1,\textbf{v}_2,\textbf{v}_3)$ linearly and gives a scalar

Sir Cumference

@Secret So...a tensor is some kind of multivariable scalar function?

Secret

It's a multilinear map. A linear map (which is a (1,1) tensor) takes in a vector and give a vector, or takes in one covector and one vector to give a scalar (e.g. in quadratic forms))

A (n,p) tensor takes in p vectors and n covectors to yield a scalar, and it is linear wrt each vector argument, thus they are like higher dimensional generalisation of matrices

@SirCumference You can also think of it as a multivariable linear function where each argument are either vectors or covectors

Sir Cumference

@Secret Ok, so what's the big deal about it?

Secret

well, take the cartesian stress tensor as an example, it gives you the stress depending on which direction the surface is facing. So if you have some physical quantity that depends linearly on multiple directional quantities, you can express that concisely with tensors

Dawood ibn Kareem

6:13 AM

@SirCumference I have a slightly woolly but kind of helpful mental model of what tensor means. It's something like "tensors = { scalars, vectors, matrices, ... and so on to any number of dimensions of writing things down}". The truth is way more complicated because you have the whole vector/covector distinction going on in each dimension; but it gives me something to have in my head when someone says "tensor". Of course, "dimension" here isn't the same as the dimension of the vector or whatever.

It's more like the number of dimensions of the paper you need to write the tensor down on.

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